If sexp had one of its fixpoints at say a + bi for 0 <a,b< oo then that would create a problem since that would require we have a straith line.
And hence sexp would then not be analytic near c + bi for any real c.

So what does that mean ?
Is not 2 a fixpoints of the half-iterate of f ?
Is there no line in this case ?

But whenever we get F(x)=2 , because of f(2)=2 we know that we get a local periodic behaviour. (period =< 1)

Guess similar question arise for most superfunctions of polynomials of degree 2 since they are all of double exponential nature.
(related : chebyshev polynomial , sinh(2^x) , mandelbrot , ... )

My confusion is probably because of the imho weird positions of solutions to F(x) = 2. ( because of ln branches ). In combination with the idea above ofcourse.

In other cases of superfunctions I can imagine the fixpoints (of the original function , not the superfunction) to be on other branches. Here that is problematic. Hence my special intrest.